User:Caliburn/s/fa/2

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Theorem

Let $\struct {X, \norm \cdot_X}$ be a Banach space.

Let $\norm \cdot$ be the norm of a bounded linear transformation.

Let $\map B {X, X}$ be the space of bounded linear transformations $X \to X$.

Let $T \in \map B {X, X}$ be a bounded linear operator with $\norm T < 1$.

Then:

$(1) \quad$ $I - T$ is invertible with inverse $\paren {I - T}^{-1}$
$(2) \quad$ $\paren {I - T}^{-1}$ is bounded.
$(3) \quad$ $\ds \paren {I - T}^{-1} = \sum_{k \mathop = 0}^\infty T^k$
$(4) \quad$ $\norm {\paren {I - T}^{-1} } \le \paren {1 - \norm T}^{-1}$


Corollary

Let $A : X \to X$ be an invertible bounded linear operator with bounded inverse $A^{-1} : X \to X$.

Let $B : X \to X$ be an invertible bounded linear operator with $\norm B \norm {A^{-1} } < 1$.


Then:

$(1) \quad$ $A + B$ is invertible with inverse $\paren {A + B}^{-1}$
$(2) \quad$ $\paren {A + B}^{-1}$ is bounded.
$(3) \quad$ $\norm {\paren {A + B}^{-1} } \le \norm {A^{-1} } \paren {1 - \norm {A^{-1} } \norm B}^{-1}$


Proof

Define the sequence $\sequence {R_n}_{n \in \N}$ by:

$\ds R_n = \sum_{k \mathop = 0}^n T^k$

for each $n \in \N$.

We first aim to show that $\sequence {R_n}_{n \in \N}$ is Cauchy.

Let $n, m$ be natural numbers with $n > m$.

Let $\epsilon$ be a positive real number.

We have:

\(\ds \norm {R_n - R_m}\) \(=\) \(\ds \norm {\sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^m T^k}\)
\(\ds \) \(=\) \(\ds \norm {\sum_{k \mathop = m + 1}^n T^k + \sum_{k \mathop = 0}^m T^k - \sum_{k \mathop = 0}^m T^k}\)
\(\ds \) \(=\) \(\ds \norm {\sum_{k \mathop = m + 1}^n T^k}\)

We have (need link):

$\norm {T^k} \le \norm T^k$

for each $k \in \N$.

So, using the triangle inequality part of Norm on Bounded Linear Transformation is Norm, we have:

\(\ds \norm {\sum_{k \mathop = m + 1}^n T^k}\) \(\le\) \(\ds \sum_{k \mathop = m + 1}^n \norm {T^k}\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = m + 1}^n \norm T^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \norm T^k - \sum_{k \mathop = 0}^m \norm T^k\)
\(\ds \) \(=\) \(\ds \frac {1 - \norm T^{n + 1} } {1 - \norm T} - \frac {1 - \norm T^{m + 1} } {1 - \norm T}\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds \frac {\norm T^{m + 1} - \norm T^{n + 1} } {1 - \norm T}\)
\(\ds \) \(\le\) \(\ds \frac {\norm T^{m + 1} } {1 - \norm T}\)

Note that we have:

$\ds \frac {\norm T^{m + 1} } {1 - \norm T} < \epsilon$

if and only if we have:

$\norm T^{m + 1} < \epsilon \paren {1 - \norm T}$

Taking logarithms, this holds if and only if:

$\paren {m + 1} \ln \norm T < \map \ln {\epsilon \paren {1 - \norm T} }$

Since:

$\norm T < 1$

we have:

$\ds m > \frac {\map \ln {\epsilon \paren {1 - \norm T} } } {\norm T} - 1$

Set:

$\ds N = \frac {\map \ln {\epsilon \paren {1 - \norm T} } } {\norm T} - 1$

Then, for $n > m > N$, we have:

$\norm {R_n - R_m} < \epsilon$

So, we have that:

$\sequence {R_n}_{n \in \N}$ is Cauchy.

From Space of Bounded Linear Transformations is Banach Space, we have that:

$\struct {\map B {X, X}, \norm \cdot}$ is a Banach space.

So:

$\sequence {R_n}_{n \in \N}$ converges in $\map B {X, X}$.

Let:

$\ds R = \lim_{n \mathop \to \infty} R_n$

We can see that:

$\ds R = \sum_{k \mathop = 0}^\infty T^k$

We now show that:

$R = \paren {I - T}^{-1}$

and that $R$ satisfies the desired properties.


Proof of $(1)$

We have:

\(\ds R_n \paren {I - T}\) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^n T^k} \paren {I - T}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\)
\(\ds \) \(=\) \(\ds I - T^{n + 1} + \sum_{k \mathop = 1}^n T^k - \sum_{k \mathop = 0}^{n - 1} T^{k + 1}\)
\(\ds \) \(=\) \(\ds I - T^{n + 1} + \sum_{k \mathop = 1}^n T^k - \sum_{k \mathop = 1}^n T^k\)
\(\ds \) \(=\) \(\ds I - T^{n + 1}\)

We have:

\(\ds \norm {R_n \paren {I - T} - I}\) \(=\) \(\ds \norm {T^{n + 1} }\)
\(\ds \) \(=\) \(\ds \norm T^{n + 1}\)
\(\ds \) \(\to\) \(\ds 0\)

so:

$\ds \lim_{n \to \infty} R_n \paren {I - T} = I$

so:

$R \paren {I - T} = I$

Similarly, we have:

\(\ds \paren {I - T} R_n\) \(=\) \(\ds \paren {I - T} \sum_{k \mathop = 0}^n T^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\)
\(\ds \) \(=\) \(\ds I - T^{n + 1}\)

So, taking $n \to \infty$ we have:

$\paren {I - T} R = I$

So $I - T$ is invertible with:

$R = \paren {I - T}^{-1}$

so we have $(1)$.

$\Box$


Proof of $(2)$

Note that since $R$ is bounded, we immediately have that:

$\paren {I - T}^{-1}$ is bounded.

$\Box$


Proof of $(3)$

Since:

$\ds R = \sum_{k \mathop = 0}^\infty T^k$

and:

$R = \paren {I - T}^{-1}$

we obtain $(3)$ immediately.

$\Box$


Proof of $(4)$

We have:

\(\ds \norm {R_n}\) \(=\) \(\ds \norm {\sum_{k \mathop = 0}^n T^k}\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 0}^n \norm {T^k}\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 0}^n \norm T^k\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 0}^\infty \norm T^k\)
\(\ds \) \(=\) \(\ds \frac 1 {1 - \norm T}\)

So, from Limits Preserve Inequalities:

$\norm R \le \dfrac 1 {1 - \norm T}$

giving the bound required by $(4)$.

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